Problem: The polar curve $r(\theta)=\theta\cdot\sin(2\theta)$ is graphed for $0\leq\theta\leq\pi$. Let $R$ be the region in the fourth quadrant enclosed by the curve and the $x$ -axis. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{\scriptsize\dfrac{\pi}{2}}^{\pi}\dfrac{1}{4}\cdot\theta^2\cdot\sin^2(2\theta)\,d\theta$ (Choice B) B $ \int^{\scriptsize\dfrac{\pi}{2}}_{0}\dfrac{1}{4}\cdot\theta^2\cdot\sin^2(2\theta)\,d\theta$ (Choice C) C $ \int^{\scriptsize\dfrac{\pi}{2}}_{0}\dfrac{1}{2}\cdot\theta^2\cdot\sin^2(2\theta)\,d\theta$ (Choice D) D $ \int_{\scriptsize\dfrac{\pi}{2}}^{\pi}\dfrac{1}{2}\cdot\theta^2\cdot\sin^2(2\theta)\,d\theta$
Explanation: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ We know $r(\theta)$ but we still need to figure out $\alpha$ and $\beta$. Notice that the curve goes through the fourth quadrant, even though it's graphed for $0\leq\theta\leq\pi$. That's because for $\theta$ -values between $\dfrac{\pi}{2}$ and $\pi$, the sign of $r(\theta)$ is negative. On that interval, the $\theta$ -values for which $r(\theta)=0$ are $0$, $\dfrac{\pi}{2}$, and $\pi$. So $\alpha$ must be $\dfrac{\pi}{2}$ and $\beta$ must be $\pi$. Let's plug ${r(\theta)=\theta\cdot\sin(2\theta)}$, ${\alpha=\dfrac{\pi}{2}}$, and ${\beta=\pi}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\scriptsize\dfrac{\pi}{2}}}^{{\pi}}\dfrac{1}{2}\left({\theta\cdot\sin(2\theta)}\right)^{2}d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{2}}^{\pi}\dfrac{1}{2}\cdot\theta^2\cdot\sin^2(2\theta)\,d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\scriptsize\dfrac{\pi}{2}}^{\pi}\dfrac{1}{2}\cdot\theta^2\cdot\sin^2(2\theta)\,d\theta$